0.25%
0.01%
0.80%
0.40%
Correct answer is C
Actual length of rope = 1.25m
Measured length of rope = 1.26m
error = (1.26 - 1.25)m - 0.01m
Percentage error = \(\frac{error}{\text{actual length}}\) x 100%
= \(\frac{0.01}{1.25}\) x 100%
= 0.8%
Simplify (\(\frac{3}{4}\) of \(\frac{4}{9}\) \(\div\) 9\(\frac{1}{2}\)) \(\div\) 1\(\frac{5}{19}\)
\(\frac{1}{5}\)
\(\frac{1}{4}\)
\(\frac{1}{36}\)
\(\frac{1}{25}\)
Correct answer is C
(\(\frac{3}{4}\) of \(\frac{4}{9}\) \(\div\) 9\(\frac{1}{2}\)) \(\div\) 1\(\frac{5}{19}\)
Applying the rule of BODMAS, we have:
(\(\frac{3}{4}\) x \(\frac{4}{9}\) \(\div\)\(\frac{19}{2}\)) \(\div\)\(\frac{24}{19}\)
(\(\frac{1}{3}\) x \(\frac{2}{19}\)) \(\div\)\(\frac{24}{19}\)
\(\frac{1}{3}\) x \(\frac{2}{19}\) x \(\frac{19}{24}\)
= \(\frac{1}{36}\)
Find r, if 6r7\(_8\) = 511\(_9\)
3
2
6
5
Correct answer is A
6r7\(_8\) = 511\(_9\)
6 x 8\(^2\) + r x 8\(^1\) + 7 x 8\(^0\) = 5 x 9\(^2\) + 1 x 9\(^1\) + 1 x 9\(^0\)
6 x 64 + 8r + 7 x 1 = 5 x 81 + 9 + 1 x 1
384 + 8r + 7 = 405 + 9 + 1
8r = 415 - 391
8r = 24
r = \(\frac{24}{8}\)
= 3
Find the derivative of \(\frac {\sin\theta}{\cos\theta}\)
sec2 \(\theta\)
tan \(\theta\) cosec \(\theta\)
cosec \(\theta\)sec \(\theta\)
cosec2\(\theta\)
Correct answer is A
\(\frac {\sin\theta}{\cos\theta}\)
\(\frac{\cos \theta {\frac{d(\sin \theta)}{d \theta}} - \sin \theta {\frac{d(\cos \theta)}{d \theta}}}{\cos^2 \theta}\)
\(\frac{\cos \theta. \cos \theta - \sin \theta (-\sin \theta)}{cos^2\theta}\)
\(\frac{cos^2\theta + \sin^2 \theta}{cos^2\theta}\)
Recall that sin2 \(\theta\) + cos2 \(\theta\) = 1
\(\frac{1}{\cos^2\theta}\) = sec2 \(\theta\)
\(\frac{2}{3}\)
\(\frac{1}{3}\)
\(\frac{2}{9}\)
\(\frac{7}{9}\)
Correct answer is C
Prime numbers = (43,47,53,59)
N = (43, 44, 45,..., 60)
The universal set contains 18 numbers.
The prime numbers between 43 and 60 are 4
Probability of picking a prime number = \(\frac{4}{18}\)
= \(\frac{2}{9}\)